Two-Digit Number and Its Reverse

Video Explanation

Question

The sum of a two-digit number and the number formed by reversing its digits is 66. If the two digits differ by 2, find the number. How many such numbers are there?

Solution

Step 1: Let the Variables

Let the tens digit = \(x\)

Let the units digit = \(y\)

Step 2: Form the Numbers

Original number = \(10x + y\)

Reversed number = \(10y + x\)

Step 3: Form the First Equation

\[ (10x + y) + (10y + x) = 66 \]

\[ 11x + 11y = 66 \]

\[ x + y = 6 \quad (1) \]

Step 4: Digits Differ by 2

Two possible cases:

Case 1: \[ x – y = 2 \quad (2) \]

Case 2: \[ y – x = 2 \quad (3) \]

Case 1: Solve (1) and (2)

x + y = 6

x – y = 2

Add the equations:

\[ 2x = 8 \]

\[ x = 4 \]

\[ y = 2 \]

Number = \(42\) —

Case 2: Solve (1) and (3)

x + y = 6

y – x = 2 \]

Add equations:

\[ 2y = 8 \]

\[ y = 4 \]

\[ x = 2 \]

Number = \(24\)

Conclusion

The required numbers are:

\[ \boxed{42 \text{ and } 24} \]

Total such numbers:

\[ \boxed{2} \]

Final Answer (For Exam)

The numbers are 42 and 24. There are 2 such numbers.